Question

Use the standard deviation to compare the variability of the datasets. Round to the nearest hundredth.

Dataset A: 4, 5, 6, 12, 13

Dataset B: 4, 5, 7, 9, 10

(1 point)
Responses

The standard deviation of Dataset A is larger than that of Dataset B, so Dataset A has more variability
The standard deviation of Dataset A is larger than that of Dataset B, so Dataset A has more variability

The standard deviation of Dataset B is larger than that of Dataset A, so Dataset B has more variability
The standard deviation of Dataset B is larger than that of Dataset A, so Dataset B has more variability

Dataset A and Dataset B have the same standard deviation, so both datasets have the same variability

Answer 1

To compare the variability of the two datasets using standard deviation, we first need to calculate the standard deviation for each dataset.

**Dataset A: 4, 5, 6, 12, 13**

1. Calculate the mean:
\[
\text{Mean}_A = \frac{4 + 5 + 6 + 12 + 13}{5} = \frac{40}{5} = 8
\]

2. Calculate the variance:
\[
\text{Variance}_A = \frac{(4-8)^2 + (5-8)^2 + (6-8)^2 + (12-8)^2 + (13-8)^2}{5}
\]
\[
= \frac{(16) + (9) + (4) + (16) + (25)}{5} = \frac{70}{5} = 14
\]

3. Calculate the standard deviation:
\[
\text{SD}_A = \sqrt{14} \approx 3.74
\]

---

**Dataset B: 4, 5, 7, 9, 10**

1. Calculate the mean:
\[
\text{Mean}_B = \frac{4 + 5 + 7 + 9 + 10}{5} = \frac{35}{5} = 7
\]

2. Calculate the variance:
\[
\text{Variance}_B = \frac{(4-7)^2 + (5-7)^2 + (7-7)^2 + (9-7)^2 + (10-7)^2}{5}
\]
\[
= \frac{(9) + (4) + (0) + (4) + (9)}{5} = \frac{26}{5} = 5.2
\]

3. Calculate the standard deviation:
\[
\text{SD}_B = \sqrt{5.2} \approx 2.28
\]

---

**Comparison:**
- Standard Deviation of Dataset A: **3.74**
- Standard Deviation of Dataset B: **2.28**

Since \( 3.74 \) is greater than \( 2.28 \), we conclude that:

**The standard deviation of Dataset A is larger than that of Dataset B, so Dataset A has more variability.**